What is the difference between the probability density f(x) and the probability P(X = x)?

For a continuous distribution, P(X = x) = 0 for any single point x. The density f(x) is the height of the curve at x; to obtain a probability you must integrate over a range, for example P(a < X < b) = \int_a^b f(x) dx.

How do I actually calculate the probability that a value falls in a range?

Integrate the density function over that range using calculus, or equivalently, convert to a Z-score with z = \frac{x - \mu}{\sigma} and look up the cumulative probability in a standard normal table. Most statistical software includes a cumulative normal function (CDF) that computes this automatically.

Can the mean \mu or standard deviation \sigma be negative?

The mean \mu can be any real number—it simply shifts the center of the distribution. However, \sigma must always be strictly positive because it measures variability or spread; a negative standard deviation is mathematically undefined for a normal distribution.

What should I do if my real data does not look perfectly normal?

Real-world data is rarely perfectly normal, but small deviations are usually acceptable; check visually with a histogram or Q-Q plot. If the data is strongly skewed, has multiple peaks, or contains extreme outliers, consider a different distribution (log-normal, exponential, etc.) or a nonparametric method.

Normal Distribution

Q: How do I actually calculate the probability that a value falls in a range?

Integrate the density function over that range using calculus, or equivalently, convert to a Z-score with z = \frac{x - \mu}{\sigma} and look up the cumulative probability in a standard normal table. Most statistical software includes a cumulative normal function (CDF) that computes this automatically.

Q: Can the mean \mu or standard deviation \sigma be negative?

The mean \mu can be any real number—it simply shifts the center of the distribution. However, \sigma must always be strictly positive because it measures variability or spread; a negative standard deviation is mathematically undefined for a normal distribution.

Q: What should I do if my real data does not look perfectly normal?

Real-world data is rarely perfectly normal, but small deviations are usually acceptable; check visually with a histogram or Q-Q plot. If the data is strongly skewed, has multiple peaks, or contains extreme outliers, consider a different distribution (log-normal, exponential, etc.) or a nonparametric method.

The normal distribution describes the probability density of a continuous variable, used to model measurements and natural phenomena.

f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}

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What each symbol means

What Normal Distribution takes

Symbol	Meaning
$f$	The probability density at a specific value of $x$ ; always non-negative and represents the height of the bell curve, but is not itself a probability.
$x$	The value of the random variable being measured (e.g., height, weight, test score); can be any real number.
$\mu$	The mean or center of the distribution, where the peak of the bell curve sits; a location parameter that shifts the entire curve left or right.
$\sigma$	The standard deviation, which controls the spread of the curve; must be positive, and larger values produce a wider, flatter bell.

When to use it

Reach for the normal distribution when you have continuous data that clusters symmetrically around a mean.

Level

Usually taught in: Statistics

Worked examples

1. Find density at one standard deviation above the mean

Problem

Find

f(1)

for a standard normal distribution with mean

\mu = 0

and standard deviation

\sigma = 1

$f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$
Start with the normal distribution formula.
$f(1) = \frac{1}{1 \cdot \sqrt{2\pi}}e^{-\frac{(1-0)^2}{2 \cdot 1^2}}$
Substitute $\mu = 0$ , $\sigma = 1$ , and $x = 1$ .
$f(1) = \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}}$
Simplify: $(1-0)^2 = 1$ and $2 \cdot 1^2 = 2$ , giving $-\frac{1}{2}$ in the exponent.
$\sqrt{2\pi} \approx 2.507$
Compute $\sqrt{2 \times 3.14159} \approx 2.507$ .
$e^{-1/2} \approx 0.6065$
Compute $e^{-0.5} \approx 0.6065$ .
$f(1) \approx \frac{0.6065}{2.507} \approx 0.242$
Divide the exponential by the denominator.

Answer: $f(1) \approx 0.242$

For a standard normal distribution, the density at one standard deviation above the mean is about 0.242. This represents the height of the bell curve at that point, not a probability—to find an actual probability, you would need to integrate over a range or use a cumulative normal table.

2. Calculate density at a value far from the mean

Problem

A manufacturing process produces parts with lengths following a normal distribution with mean

\mu = 100

mm and standard deviation

\sigma = 15

mm. Find

f(130)

, the probability density at 130 mm.

$f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$
Begin with the normal distribution formula.
$f(130) = \frac{1}{15\sqrt{2\pi}}e^{-\frac{(130-100)^2}{2 \cdot 15^2}}$
Substitute $x = 130$ , $\mu = 100$ , and $\sigma = 15$ .
$x - \mu = 130 - 100 = 30$
Calculate the deviation from the mean.
$(x-\mu)^2 = 30^2 = 900$
Square the deviation.
$2\sigma^2 = 2 \cdot 15^2 = 2 \cdot 225 = 450$
Compute twice the squared standard deviation.
$-\frac{900}{450} = -2$
Calculate the exponent: $-\frac{(x-\mu)^2}{2\sigma^2} = -2$ .
$\sigma\sqrt{2\pi} = 15 \times 2.507 \approx 37.605$
Compute the denominator.
$e^{-2} \approx 0.1353$
Evaluate the exponential.
$f(130) \approx \frac{0.1353}{37.605} \approx 0.0036$
Divide to obtain the final density.

Answer: $f(130) \approx 0.0036$

At a distance of 2 standard deviations above the mean, the density is much smaller than at the peak. This is characteristic of the bell curve: values far from the center are far less likely, and the density tapers off. This density value still does not represent a probability directly; integration is needed for that.

3. Word problem: cookie weight at a school bake sale

Problem

At a school fundraiser bake sale, the weight of chocolate chip cookies follows a normal distribution with mean

\mu = 45

g and standard deviation

\sigma = 4

g. Find the probability density

f(50)

at a cookie weighing 50 g.

$f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$
Start with the normal distribution probability density function.
$f(50) = \frac{1}{4\sqrt{2\pi}}e^{-\frac{(50-45)^2}{2 \cdot 4^2}}$
Substitute $x = 50$ , $\mu = 45$ , and $\sigma = 4$ grams.
$50 - 45 = 5$
Calculate how far 50 g is from the mean.
$(50-45)^2 = 5^2 = 25$
Square the deviation.
$2\sigma^2 = 2 \cdot 4^2 = 2 \cdot 16 = 32$
Compute twice the squared standard deviation.
$-\frac{25}{32} \approx -0.78125$
Calculate the exponent as a decimal.
$4\sqrt{2\pi} \approx 4 \times 2.507 \approx 10.028$
Compute the denominator.
$e^{-0.78125} \approx 0.4573$
Evaluate the exponential term.
$f(50) \approx \frac{0.4573}{10.028} \approx 0.0456$
Divide the exponential by the denominator.

Answer: $f(50) \approx 0.0456$

A cookie weighing 50 g is about 1.25 standard deviations above the mean (since $\frac{50-45}{4} = 1.25$ ). The density at this weight is still fairly high, reflecting that 50 g is not too far from the average 45 g. The density alone does not tell us how many cookies weigh exactly 50 g or what proportion fall in a range; for that, integrate over the desired interval.

Common mistakes

Where Normal Distribution usually goes wrong

The mistake	Why it is wrong	The fix
Treating $f(50) = 0.0456$ as a probability, saying 'there is a 4.56% chance the cookie weighs exactly 50 g.'	For any continuous distribution, $P(X = x) = 0$ for a single exact value; $f(x)$ is a probability density, not a probability itself.	To find a probability, integrate the density over a range (e.g., $P(48 < X < 52) = \int_{48}^{52} f(x) dx$ ) or use a Z-score table with cumulative probabilities.
Using a negative standard deviation, such as $\sigma = -4$ , without catching the error.	Standard deviation must be positive because it measures the spread or distance of data from the mean; negative spread is nonsensical and produces an invalid distribution.	Always ensure $\sigma > 0$ . If a calculation yields a negative value, recheck your formula; if data is given, verify that you computed standard deviation correctly.
Forgetting to subtract the mean and computing $-\frac{x^2}{2\sigma^2}$ instead of $-\frac{(x-\mu)^2}{2\sigma^2}$ .	The normal distribution is centered at $\mu$ , not at zero; omitting the subtraction shifts where the peak occurs and produces incorrect densities at every point.	Always compute $(x - \mu)$ first before squaring. For example, if $x = 50$ and $\mu = 45$ , calculate $(50 - 45) = 5$ , then $(5)^2 = 25$ .

Tips and when to use something else

The area under the entire curve equals 1, but $f(x)$ itself can exceed 1—never confuse density with probability.
For practical probability calculations, convert to Z-scores using $z = \frac{x - \mu}{\sigma}$ and use a standard normal table instead of computing integrals by hand.
Use the Empirical Rule for a quick check: roughly 68% of data falls within 1 $\sigma$ of $\mu$ , 95% within 2 $\sigma$ , and 99.7% within 3 $\sigma$ .
Always verify that your data is approximately normal before using this formula; check a histogram or Q-Q plot first, especially if the data is skewed or has obvious outliers.

Related formulas

Standard DeviationStandard deviation measures how spread out data points are from the average—use it to understand variability in a dataset.Z-ScoreThe Z-Score tells you how many standard deviations a data point is from the mean, revealing whether it is typical or unusual.Empirical RuleThe Empirical Rule describes where approximately 68%, 95%, and 99.7% of normally distributed data fall, measured in standard deviations from the mean.Confidence Interval for a MeanEstimates the range where a population mean likely falls using sample data when the population standard deviation is unknown.Standard Error of the MeanThe standard error of the mean measures how much sample means vary around the true population mean; use it for confidence intervals and hypothesis tests.

Frequently asked questions

What is the difference between the probability density $f(x)$ and the probability $P(X = x)$ ?: For a continuous distribution, $P(X = x) = 0$ for any single point $x$ . The density $f(x)$ is the height of the curve at $x$ ; to obtain a probability you must integrate over a range, for example $P(a < X < b) = \int_a^b f(x) dx$ .
How do I actually calculate the probability that a value falls in a range?: Integrate the density function over that range using calculus, or equivalently, convert to a Z-score with $z = \frac{x - \mu}{\sigma}$ and look up the cumulative probability in a standard normal table. Most statistical software includes a cumulative normal function (CDF) that computes this automatically.
Can the mean $\mu$ or standard deviation $\sigma$ be negative?: The mean $\mu$ can be any real number—it simply shifts the center of the distribution. However, $\sigma$ must always be strictly positive because it measures variability or spread; a negative standard deviation is mathematically undefined for a normal distribution.
What should I do if my real data does not look perfectly normal?: Real-world data is rarely perfectly normal, but small deviations are usually acceptable; check visually with a histogram or Q-Q plot. If the data is strongly skewed, has multiple peaks, or contains extreme outliers, consider a different distribution (log-normal, exponential, etc.) or a nonparametric method.

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Reviewed 2026-09-18